Zero-initializing elements of a std::array with a default member initializer
Suppose I have a class template like this:
template<typename T, size_t N>
struct S {
std::array<T,N> a;
};
Is there a default member initializer I can place on a,
template<typename T, size_t N>
struct S {
std::array<T,N> a = ???;
};
such that no matter what T is, the elements of a will always be initialized (never have indeterminant value)? I.e., even if T is a primitive type like int.
c++ c++17
edited yesterday
Peter Mortensen
13.5k1983111
asked yesterday
Andrew Tomazos
34.6k25132227
add a comment |
Suppose I have a class template like this:
template<typename T, size_t N>
struct S {
std::array<T,N> a;
};
Is there a default member initializer I can place on a,
template<typename T, size_t N>
struct S {
std::array<T,N> a = ???;
};
such that no matter what T is, the elements of a will always be initialized (never have indeterminant value)? I.e., even if T is a primitive type like int.
c++ c++17
edited yesterday
Peter Mortensen
13.5k1983111
asked yesterday
Andrew Tomazos
34.6k25132227
add a comment |
Suppose I have a class template like this:
template<typename T, size_t N>
struct S {
std::array<T,N> a;
};
Is there a default member initializer I can place on a,
template<typename T, size_t N>
struct S {
std::array<T,N> a = ???;
};
such that no matter what T is, the elements of a will always be initialized (never have indeterminant value)? I.e., even if T is a primitive type like int.
c++ c++17
edited yesterday
Peter Mortensen
13.5k1983111
asked yesterday
Andrew Tomazos
34.6k25132227
Suppose I have a class template like this:
template<typename T, size_t N>
struct S {
std::array<T,N> a;
};
Is there a default member initializer I can place on a,
template<typename T, size_t N>
struct S {
std::array<T,N> a = ???;
};
such that no matter what T is, the elements of a will always be initialized (never have indeterminant value)? I.e., even if T is a primitive type like int.
c++ c++17
c++ c++17
edited yesterday
Peter Mortensen
13.5k1983111
asked yesterday
Andrew Tomazos
34.6k25132227
edited yesterday
Peter Mortensen
13.5k1983111
asked yesterday
Andrew Tomazos
34.6k25132227
edited yesterday
Peter Mortensen
13.5k1983111
edited yesterday
Peter Mortensen
13.5k1983111
edited yesterday
Peter Mortensen
13.5k1983111
13.5k1983111
asked yesterday
Andrew Tomazos
34.6k25132227
asked yesterday
Andrew Tomazos
34.6k25132227
asked yesterday
Andrew Tomazos
34.6k25132227
34.6k25132227
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
This:
template<typename T, size_t N>
struct S {
std::array<T,N> a = {};
};
That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:
struct A {
A() {}
int i;
};
which would prevent i from being initialized. But that's on them.
answered yesterday
Barry
177k18304561
add a comment |
std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.
answered yesterday
Jans
8,51822635
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
This:
template<typename T, size_t N>
struct S {
std::array<T,N> a = {};
};
That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:
struct A {
A() {}
int i;
};
which would prevent i from being initialized. But that's on them.
answered yesterday
Barry
177k18304561
add a comment |
This:
template<typename T, size_t N>
struct S {
std::array<T,N> a = {};
};
That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:
struct A {
A() {}
int i;
};
which would prevent i from being initialized. But that's on them.
answered yesterday
Barry
177k18304561
add a comment |
This:
template<typename T, size_t N>
struct S {
std::array<T,N> a = {};
};
That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:
struct A {
A() {}
int i;
};
which would prevent i from being initialized. But that's on them.
answered yesterday
Barry
177k18304561
This:
template<typename T, size_t N>
struct S {
std::array<T,N> a = {};
};
That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:
struct A {
A() {}
int i;
};
which would prevent i from being initialized. But that's on them.
answered yesterday
Barry
177k18304561
edited yesterday
answered yesterday
Barry
177k18304561
answered yesterday
Barry
177k18304561
answered yesterday
Barry
177k18304561
177k18304561
add a comment |
add a comment |
std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.
answered yesterday
Jans
8,51822635
add a comment |
std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.
answered yesterday
Jans
8,51822635
add a comment |
std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.
answered yesterday
Jans
8,51822635
std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.
answered yesterday
Jans
8,51822635
edited yesterday
answered yesterday
Jans
8,51822635
answered yesterday
Jans
8,51822635
answered yesterday
Jans
8,51822635
8,51822635
add a comment |
add a comment |
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